Question

If In triangle ABC a =9, b = 8, and c=4,prove that cos b - 2 cos c= -4/3

Answer 1

Answer:

$cosB-2\:cosC=\frac{-4}{3}$

Step-by-step explanation:

I have applied cosine formula to solve this problem

$\text{Cosine formula:}$

$\boxed{cosB=\frac{c^2+a^2-b^2}{2ca}\:and\:cosC=\frac{a^2+b^2-c^2}{2ab}}$

$cosB-2\:cosC$

$=\frac{c^+a^2-b^2}{2ca}-2(\frac{a^2+b^2-c^2}{2ab})$

$=\frac{4^2+9^2-8^2}{2(4)(9)}-2(\frac{9^2+8^2-4^2}{2(9)(8)})$

$=\frac{16+81-64}{72}-\frac{81+64-16}{72}$

$=\frac{33}{72}-\frac{129}{72}$

$=\frac{-96}{72}$

$=\frac{-4}{3}$

$\implies\:\boxed{cosB-2\:cosC=\frac{-4}{3}}$